SOLUTION: I had a problem I was working on with the quadratic equation rule. {{{x^2+4x+2=0}}} Eventually I worked the problem down to {{{x= -2 +or- radical 8}}} The answer I'm suppo

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Question 68138: I had a problem I was working on with the quadratic equation rule.
x%5E2%2B4x%2B2=0
Eventually I worked the problem down to x=+-2+%2Bor-+radical+8
The answer I'm suppose to get from there is -2+%2Bor-+radical+2
I don't see how this is possible.
Found 2 solutions by stanbon, funmath:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+4x+2=0
a=1, b=4, c=2
x=[-4+-sqrt(4^2-4*2]/2
x=[-4+-sqrt8]/2
x=[-4+-2sqrt2]/2
Dividing by "2" you get:
x=[-2+-sqrt2]
Cheers,
Stan H.

Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
I had a problem I was working on with the quadratic equation rule.
This is how you solve it by completing the square.
x%5E2%2B4x%2B2=0
x%5E2%2B4x%2B2-2=0-2
x%5E2%2B4x=-2
x%5E2%2B4x%2B4=4-2
x%5E2%2B4x%2B4=2
%28x%2B2%29%5E2=2
sqrt%28%28x%2B2%29%5E2%29=%2B-sqrt%282%29
x%2B2=%2B-sqrt%282%29
x%2B2-2=-2%2B-sqrt%282%29
highlight%28x=-2%2B-sqrt%282%29%29
:
If you were supposed to solve it using the quadratic formula:
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
a=1, b=4 and c=2
x=%28-4%2B-sqrt%28%284%29%5E2-4%281%29%282%29%29%29%2F%282%281%29%29
x=%28-4%2B-sqrt%2816-8%29%29%2F2
x=%28-4%2B-sqrt%288%29%29%2F2
x=%28-4%2B-sqrt%284%29%2Asqrt%282%29%29%2F2
x=%28-4%2Bor-2%2Asqrt%282%29%29%2F2
x=-4%2F2%2Bor-2%2Asqrt%282%29%2F2
highlight%28x=-2%2B-sqrt%282%29%29
Either way you eventually get there.
Happy Calculating!!!!